Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 0 - Before Calculus - 0.4 Inverse Functions - Exercises Set 0.4 - Page 45: 20

Answer

$f^{-1}(x) = \frac{1+\sqrt {1-20x}}{10} $, where $x\leq -4$

Work Step by Step

We first write, $x= f(y) =y − 5y^2, x \geq 1\Rightarrow y\leq -4$ Then we solve this equation for $x$ as a function of $y$ $x=y-5y^2$ or $-x-5y^2+y=0$ or $5y^2-y+x=0$ For solving $y$ use the Quadratic formula: $y = \frac{1\pm\sqrt {1-20x}}{10} $ where $x\leq -4$ Since $y\geq 1$ we have the inverse: $f^{-1}(x) = \frac{1+\sqrt {1-20x}}{10} $ where $x\leq -4$
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